Question

plz answer this question its really urgent

Answer 1

LHL at x=0

[tex]\lim_{h \to 0} f(0-h) = \lim_{h \to 0} [1+3(0-h)]^{1/(0-h)}    = \lim_{h \to 0} [1-3h]^{-1/h} = \lim_{h \to 0} e^{ln(1-3h)^{-1/h}[/tex][tex]\lim_{h \to 0} e^{(-1/h)ln(1-3h) = e^{ \lim_{h \to 0} \frac{ln(3h-1)}{h}[/tex]

By using L hopital's

[tex]e^ \lim_{h \to 0} \frac{3}{3h-1}   =e^{3/-1} =e^-3[/tex]

By observation RHL =e^3/1 = e^3

so LHL ≠RHL